Question

In given figure DE||BC and $AD:DB=5:4$ Find $\frac{Area\left(△DEF\right)}{Area\left(△CFB\right)}$.

Answer 1

In $△ABC$, we have

DE||BC

$\Rightarrow$ $\angle ADE=\angle ABC$ and $\angle AED=\angle ACB$ [Corresponding angles]

Thus, in triangles ADE and ABC, we have

$\angle A=\angle A$ [Common]

$\angle ADE=\angle ABC$

and, $\angle AED=\angle ACB$

$\therefore$ $△AED\sim △ABC$ [By AAA similarity]

$\Rightarrow$ $\frac{AD}{AB}=\frac{DE}{BC}$

We have,

$\frac{AD}{DB}=\frac{5}{4}$

$\Rightarrow$ $\frac{DB}{AD}=\frac{4}{5}$

$\Rightarrow$ $\frac{DB}{AD}+1=\frac{4}{5}+1$

$\Rightarrow$ $\frac{DB+AD}{AD}=\frac{9}{5}$

$\Rightarrow$ $\frac{AB}{AD}=\frac{9}{5}\Rightarrow \frac{AD}{AB}=\frac{5}{9}$

$\therefore$ $\frac{DE}{BC}=\frac{5}{9}$

In $△DFE$ and $△CFB$, we have

$\angle 1=\angle 3$ [Alternate interior angles]

$\angle 2=\angle 4$ [Vertically opposite angles]

Therefore, by AA-similarity criterion, we have

$△DFE\sim △CFB$

$\Rightarrow$ $\frac{Area\left(△DFE\right)}{Area\left(△CFB\right)}=\frac{{DE}^2}{{BC}^2}$

$\Rightarrow$ $\frac{Area\left(△DFE\right)}{Area\left(△CFB\right)}={\left(\frac{5}{9}\right)}^2=\frac{25}{81}$.....[Using (i)]